Question

Let g(x)=x^3+12x^2+36xg(x)=x 3 +12x 2 +36xg, left parenthesis, x, right parenthesis, equals, x, start superscript, 3, end superscript, plus, 12, x, start superscript, 2, end superscript, plus, 36, x and let ccc be the number that satisfies the Mean Value Theorem for ggg on the interval -8\leq x\leq-2−8≤x≤−2minus, 8, is less than or equal to, x, is less than or equal to, minus, 2.

Answer 1

Answer with Step-by-step explanation:

Mean value theorem:If a function f(x) is continuous on [a,b] and differentiable on (a,b).Then there exist a number $c\in(a,b)$such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

$f(x)=x^3+12x^2+36x$

We are given that a=-8 and b=-2

$f(-8)=(-8)^3+12(-8)^2+36(-8)=-512+768-288=-32$

$f(-2)=(-2)^2+12(-2)^2+36(-2)=8+48-72=-16$

$f'(x)=3x^2+24x+36$

substitute x=c then we get

$f'(c)=3c^2+24c+36$

Substitute the values in the formula

$3c^2+24c+36=\frac{-16+32}{-2+8}$

$3c^2+24c+36=\frac{16}{6}$

$3c^2+24c+36=\frac{8}{3}$

$9c^2+72c+108=8$

$9c^2+72c+108-8=0$

$9c^2+72c+100=0$

Quadratic formula for $ax^2+bx+c$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Using the quadratic formula

$c=\frac{-72\pm\sqrt{5184-3600}}{2\times 9}$

$c=\frac{-72\pm\sqrt{1584}}{18}$

$c=\frac{-72\pm 39.799}{18}$

$c=\frac{-72+39.799}{18}$and $c=\frac{-72-39.799}{18}$

$c=-1.789$ and $c=-6.21$

c=-1.789 does not lie in the interval [-8,-2]

But c=-6.21 lies in the interval

Hence, the function satisfied the mean value theorem.